## RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17C.

**Other Exercises**

- RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17A
- RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17B
- RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17C
- RS Aggarwal Solutions Class 7 Chapter 17 Constructions CCE Test Paper

**Mark (✓) against the correct answer in each of the following:**

**Question 1.**

**Solution:**

(c) Supplement of 45° is 135°

135°+ 45° = 180°

**Question 2.**

**Solution:**

(b) Complement of 80° is 10°

10° + 80° = 90°

**Question 3.**

**Solution:**

(b) The angle is its own complement.

The measure of the angles will be 45° (45° + 45° = 90°)

**Question 4.**

**Solution:**

(a) The angle is one-fifth of its supplement

Let angle be x, then

x + 5x = 180°

⇒ 6x = 180°

⇒ x = 30°

Angle is 30°

**Question 5.**

**Solution:**

(b) Let angle is x

Then its complement angle=x-24° But x + x- 24° = 90°

⇒ 2x = 90° + 24° = 114°

⇒ x = 57°

The required angle is 57°

**Question 6.**

**Solution:**

(b) Let required angle = x

Then its supplement angle = x + 32

But x + x + 32° = 180°

⇒ 2x = 180° – 32 = 148°

⇒ x = 74°

Required angle = 74°

**Question 7.**

**Solution:**

(c) Two supplementary angle are in the ratio = 3 : 2

Let first angle = 3x

Second angle = 2x

But 3x + 2x = 180°

⇒ 5x = 180°

⇒ x = 36°

Smaller angle = 2x = 2 x 36° = 72°

**Question 8.**

**Solution:**

(b) In the figure ∠BOC = 132°

But ∠AOC + ∠BOC =180° (Linear pair)

⇒ ∠AOC + 132° = 180°

⇒ ∠AOC = 180° – 132° = 48°

**Question 9.**

**Solution:**

(c) In the figure, ∠AOC = 68°

But ∠AOC + ∠BOC = 180° (Linear pair)

⇒ 68° + x = 180°

⇒ x = 180° – 68° = 112°

**Question 10.**

**Solution:**

(b) In the figure,

AOB is a straight line

∠AOC + ∠BOC = 180° (Linear pair)

⇒ 2x – 10° + 3x + 15° = 180°

⇒ 5x = 180° + 10° – 15° = 175°

⇒ x = 35°

x = 35

**Question 11.**

**Solution:**

(d) In the figure,

AOB is a straight line

∠AOC + ∠COD + ∠DOB = 180°

⇒ 55° + x + 45° = 180°

⇒ x + 100° = 180°

⇒ x = 180° – 100° = 80°

**Question 12.**

**Solution:**

(a) AOB is a straight line

x + y = 180°

But 4x = 5y

**Question 13.**

**Solution:**

(b) AB and CD intersect each other at O and ∠AOC = 50°

∠BOD = ∠AOC = 50° (Vertically opposite angles)

**Question 14.**

**Solution:**

(a) AOB is a straight line

∠AOC + ∠COD + ∠DOB = 180°

⇒ 3x – 8° + 50° + x + 10° = 180°

⇒ 4x = 180° + 8° – 50° – 10°

⇒ 4x = 128°

⇒ x = 32°

**Question 15.**

**Solution:**

(b) In ∆ABC, side BC is produced to D

∠ACD = 132° and ∠A = 54°

Ext. ∠ACD = ∠A + ∠B

⇒ 132° = 54° + ∠B

⇒ ∠B = 132° – 54° = 78°

**Question 16.**

**Solution:**

(c) In ∆ABC,

Side BC is produced to D

∠A = 45°, ∠B = 55°

Ext. ∠ACD = ∠A + ∠B = 45° + 55° = 100°

**Question 17.**

**Solution:**

(b) In ∆ABC, side BC is produced to D

∠ABC = 70° and ∠ACD = 120°

Ext. ∠ACD = ∠BAC + ∠ABC

⇒ 120° = ∠BAC + 70°

⇒ ∠BAC = 120° – 70° = 50°

**Question 18.**

**Solution:**

(c) In the figure,

∠AOB = 50°, ∠BOC = 90°

∠COD = 70°, ∠AOD = x.

But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360° (Angles at a point)

⇒ 50° + 90° + 70° + x = 360°

⇒ 210 + x = 360°

⇒ x = 360° – 210°

⇒ x = 150°

**Question 19.**

**Solution:**

(c) In the figure,

Side BC of ∆ABC is produced to D

CE || BA is drawn

∠A = 50° and ∠ECD = 60°

AB || CE

∠ABC = ∠ECD (corresponding angle) = 60°

But in ∆ABC,

∠A + ∠B + ∠ACB = 180° (Angles of a triangles)

⇒ 50° + 60° + ∠ACB = 180°

⇒ ∠ACB = 180° – 50° – 60° = 70°

**Question 20.**

**Solution:**

(b) In ∆ABC,

∠A = 65°, ∠C = 85°

But ∠A + ∠B + ∠C = 180° (Angles of a triangle)

⇒ 65° +∠B+ 85° = 180°

⇒ 150° + ∠B = 180°

⇒ ∠B = 180° – 150° = 30°

**Question 21.**

**Solution:**

(d) Sum of angles of a triangle = 180°

**Question 22.**

**Solution:**

(c) Sum of angles of a quadrilateral = 360°

**Question 23.**

**Solution:**

(b) In the figure, AB || CD

∠OAB = 150°, ∠OCD = 120°

From O, draw OE || AB or CD

AB || DE

∠OAB + ∠AOE =180°

⇒ 150° + ∠AOE = 180°

⇒ ∠AOE = 180° – 150° = 30°

Similarly DE || CD

∠EOC + ∠OCD = 180°

⇒ ∠EOC + 120° = 180°

⇒ ∠EOC = 180° – 120° = 60°

Now ∠AOC = ∠AOE + ∠EOC = 30° + 60° = 90°

**Question 24.
**

**Solution:**

(a) In the given figure,

PQ || RS,

∠PAB = 60° and ∠ACS = 100°

PQ || RS

∠ABC = ∠PAB (alternate angles) = 60°

But Ext. ∠ACS = ∠BAC + ∠ABC

⇒ 100° = ∠BAC + 60°

⇒ ∠BAC = 100° – 60° = 40°

**Question 25.**

**Solution:**

(c) In the figure, AB || CD || EF

∠ABG =110° and ∠GCD = 100°

∠BGC = x°

AB || EF

∠ABG + ∠BGE = 180°

⇒ 110° + ∠BGE = 180°

⇒ ∠BGE = 180° – 110° = 70°

Similarly CD || EF

∠GCD + ∠CGF = 180°

⇒ 100° + ∠CGF = 180°

⇒ ∠CGF = 180° – 100° = 80°

But ∠BGE + ∠BGC + ∠CGF = 180°

⇒ 70° + x + 80° = 180°

⇒ 150° + x = 180°

⇒ x = 180° – 150° = 30°

**Question 26.**

**Solution:**

(d) Sum of any two sides of a triangle is always greater than the third side

**Question 27.**

**Solution:**

(d) The diagonals of a rhombus always bisect each other at right angles.

**Question 28.**

**Solution:**

(c) In ∆ABC, ∠B = 90°

AB = 5 cm and AC = 13 cm

But AC² = AB² + BC² (By Pythagoras Theorem)

⇒ (13)² = (5)² + BC²

⇒ 169 = 25 + BC2

⇒ BC² = 169 – 25 = 144 = (12)²

BC = 12 cm

**Question 29.**

**Solution:**

(c) In ∆ABC, ∠B = 37°, ∠G = 29°

But ∠A + ∠B + ∠C = 180° (angles of a triangle)

⇒ ∠A + 37° + 29° = 180°

⇒ ∠A + 66° = 180°

⇒ ∠A = 180° – 66° = 114°

**Question 30.**

**Solution:**

(c) The ratio of angles of a triangle is 2 : 3 : 7

But sum of angles of a triangle = 180°

**Question 31.**

**Solution:**

In ∆ABC,

Let 2∠A = 3∠B = 6∠C = x

**Question 32.**

**Solution:**

(a) In ∆ABC,

∠A + ∠B = 65°, ∠B + ∠C = 140°

∠A = 65°

∠C = 140° – ∠B

But ∠A + ∠B + ∠C = 180° (Angles of a triangle)

⇒ 65° – ∠B + 140° – ∠B + ∠B = 180°

⇒ 205° – ∠B = 180°

⇒ ∠B = 205° – 180° = 25°

**Question 33.**

**Solution:**

(b) In ∆ABC, ∠A – ∠B = 33°

and ∠B – ∠C = 18°

∠A = 33° + ∠B and ∠C = ∠B – 18°

But ∠A + ∠B + ∠C = 180°

⇒ 33° + ∠B + ∠B + ∠B – 18° = 180°

⇒ 3∠B = 180° – 33° + 18° = 165°

⇒ ∠B = 55°

**Question 34.**

**Solution:**

(c) In ∆ABC

∠A + ∠B + ∠C= 180° (Sum of angles of a triangle)

But angles are (3x)°, (2x – 7)° and (4x – 11)°

3x + (2x – 7) + (4x – 11)° = 180°

⇒ 3x + 2x – 7 + 4x – 11° = 180°

⇒ 9x – 18° = 180°

⇒ 9x = 180° + 18° = 198°

⇒ x = 22°

**Question 35.**

**Solution:**

(c) ∆ABC is a right angled, ∠A = 90°

AB = 24 cm, AC = 7 cm

but BC² = AB² + AC²

⇒ BC² = (24)² + (7)² = 576 + 49 = 625 = (25)²

BC = 25 cm

**Question 36.**

**Solution:**

(b) Let AB is a ladder and A is the window

BC = 15 m, AC = 20 m

Now in right ∆ABC

AB² = BC² + AC² = (15)² + (20)² = 225 + 400 = 625 = (25)²

AB = 25 m

Length of ladder = 25 m

**Question 37.**

**Solution:**

(a) Let AB and CD are two poles such that

AB = 6 m, CD = 11 m

and distance between two poles BD = 12m

From A, draw AE || BD

AE = BD = 12m

CE = CD – ED = 11 – 6 = 5 m

Now in right ∆AEC

AC² = AE² + CE² = (12)² + (5)² = 144 + 25 = 169 = (13)²

AC = 13 m

Distance between tops of poles = 13 m

**Question 38.**

**Solution:**

(d) ∆ABC is an isosceles triangle

∠C = 90°,

AC = 5 cm

BC = AC = 5 cm

In right ∆ABC

AB² = AC² + BC² = (5)² + (5)² = 25 + 25 = 50 = 2 x 25

AB = √(2 x 25) = 5√2 cm

Hope given RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17C are helpful to complete your math homework.

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